106 Construct Binary Tree from Inorder and Postorder Traversal-优快云博客

本文介绍了一种通过给定的中序遍历和后序遍历序列来构建二叉树的方法。利用递归策略，该算法能够准确地还原原始二叉树结构。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTree(inorder, 0, inorder.length,
                        postorder, 0, postorder.length);
        
    }
    
    private TreeNode buildTree(int[] inorder, int b1, int e1,
                              int[] postorder, int b2, int e2) {
        if (b1 == e1)
            return null;
        if (b2 == e2)
            return null;
        
        TreeNode root = new TreeNode(postorder[e2 - 1]);
        int inPos = find(inorder, b1, e1, root.val);
        int leftSize = inPos - b1;
        
        root.left = buildTree(inorder, b1, inPos,
                             postorder, b2, b2 + leftSize);
        root.right = buildTree(inorder, inPos + 1, e1,
                              postorder, b2 + leftSize, e2 - 1);
        
        return root;
    }
    
    private int find(int[] array, int b, int e, int val) {
        for (int i = b; i < e; i++) {
            if (array[i] == val)
                return i;
        }
        
        return -1;
    }
}